(2t-1)(t+5)=-9

Solution for (2t-1)(t+5)=-9 equation:

(2t-1)(t+5)=-9

We move all terms to the left:

(2t-1)(t+5)-(-9)=0

We add all the numbers together, and all the variables

(2t-1)(t+5)+9=0

We multiply parentheses ..

(+2t^2+10t-1t-5)+9=0

We get rid of parentheses

2t^2+10t-1t-5+9=0

We add all the numbers together, and all the variables

2t^2+9t+4=0

a = 2; b = 9; c = +4;
Δ = b2-4ac
Δ = 92-4·2·4
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-7}{2*2}=\frac{-16}{4}
=-4
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+7}{2*2}=\frac{-2}{4}
=-1/2
$